nLab amnestic isofibration

Redirected from "amnestic isofibrations".

Contents

Idea

A functor is an amnestic isofibration if it is an amnestic functor and an isofibration. The combination of these properties admits a particularly simply description in terms of a lifting property.

Proposition

A functor U:𝒟𝒞U : \mathcal{D} \to \mathcal{C} is an amnestic isofibration if and only if, for any object DD in 𝒟\mathcal{D} and any isomorphism f:CUDf : C \to U D in 𝒞\mathcal{C}, there is a unique isomorphism f˜:C˜D\tilde{f} : \tilde{C} \to D such that Uf˜=fU \tilde{f} = f.

Proof

Indeed, if f˜:C˜D\tilde{f}' : \tilde{C}' \to D were any other isomorphism such that Uf˜=fU \tilde{f}' = f, then U(f˜ 1f˜)=id CU (\tilde{f}^{-1} \circ \tilde{f}') = id_C, so we must have f˜=f˜\tilde{f} = \tilde{f}'.

Amnestic isofibrations are occasionally called discrete isofibrations, but this term may be misleading, because they are not isofibrations with discrete fibres.

Properties

Lemma

An equivalence of categories is an isomorphism if and only if it is an amnestic isofibration.

Proof

Every equivalence is essentially surjective on objects. If it is an isofibration, it is actually surjective on objects. If it is fully faithful and amnestic, then it is injective on objects. Under the given assumptions, the equivalence is thus surjective and injective on objects, and full faithfulness implies bijectivity on morphisms, hence it is an isomorphism.

Proposition

A monadic functor is strictly monadic if and only if it is also an amnestic isofibration.

Proof

Clearly, a strictly monadic functor is an amnestic isofibration; and if a monadic functor UU is amnestic, then the comparison functor KK is also amnestic, and if UU is a monadic isofibration, so is KK; therefore in this case KK must be an isomorphism of categories.

Examples

References

A reference using the term amnestic isofibrations:

A reference using the term discrete isofibrations:

Last revised on July 31, 2026 at 14:33:03. See the history of this page for a list of all contributions to it.